Question

Notify Me Bookmark

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields: electric field E→ and magnetic field B→. The beam then enters a region of another magnetic field, B′→, which is perpendicular to the ion beam. The radius of curvature of the resulting ion beam will be proportional to what? B/EB′ BB′/E E/BB′ EB′/B EB/B′

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields: electric field E→ and magnetic field B→. The beam then enters a region of another magnetic field, B′→, which is perpendicular to the ion beam. The radius of curvature of the resulting ion beam will be proportional to what? B/EB′ BB′/E E/BB′ EB′/B EB/B′

Image text
In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields: electric field E → and magnetic field B → . The beam then enters a region of another magnetic field, B ′ → , which is perpendicular to the ion beam. The radius of curvature of the resulting ion beam will be proportional to what? B / E B ′ B B ′ / E E / B B ′ E B ′ / B E B / B ′

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

A scientist builds an apparatus to measure the charge to mass ratio of ions. An ion is first accelerated to a speed of 4.50×107 m/s. It then passes through a region of magnetic field, with a magnitude of 2.44×10−2 T, perpendicular to the ion's velocity. Because the ion is moving so fast, it is only in the presence of the magnetic field for 2.00×10−7 s. Upon exiting the magnetic field, the scientist measures that the ion was deflected a distance of 3.00 cm in a direction perpendicular to its initial velocity. (a) What is the ratio of the absolute value of the ion's charge to its mass (in C/kg )? Assume the velocity change is small, and that the component of the velocity along the ion's original direction does not change noticeably. |q| m = Cg (b) Suppose the scientist determines the ion is singly charged; that is, the magnitude of its charge is that of a proton. What is the mass of the ion (in kg)? kg

Solution
0 0

J. J. Thomson is best known for his discoveries about the nature of cathode rays. His other important contribution was the invention, together with one of his students, of the mass spectrometer, a device that measures the ratio of mass m to (positive) charge q of an ion. The spectrometer consists of two regions as shown in the figure. (Figure 1) In the first region an electric field accelerates the ion and in the second the ion follows a circular arc in a magnetic field. The radius of curvature of the arc can be measured and then the m/q ratio can be found. Figure 1 of 1 After being accelerated to a speed of 1.69×105 m/s, the particle enters a uniform magnetic field of strength 0.900 T and travels in a circle of radius 35.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q. Find the ratio m/q for this particle. Express your answer numerically in kilograms per coulomb. View Available Hint(s)
Solution
0 0

A positively charged particle of mass m and charge q is released from rest in a uniform electric field, E→ = (10i^) V/m. The particle, after travelling 2 s in this electric field, enters a region that is free from electric field but has a steady and uniform magnetic field, B→ = (−4k^) T. What will be the radius of the circular path (in meters) and the direction of angular velocity of the particle in this region? (Select the closest number among the following choices.) (Note: an incorrect answer will be penalized 2 pts.) a) 5.0; k^ b) 0.2; k^ c) 0.2; −k^ d) 2.0; k^ e) 5.0; −k^
Solution
0 0